A cyclotron's oscillator frequency is 10MHz. What should be the operating magnetic field for accelerating protons ? If the radius of it's dees is 60cm what is the kinetic energy ( in MeV) of the proton beam produced by the accelerator?
[ e = 1.60×10^-19C , mp = 1.67×10^-27Kg , 1MeV = 1.6×10^-13 J ]
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Magnetic field ... Cyclotron's oscillator frequency should be same as the proton's revolution frequency .
[tex]f \: = \frac{ Beq }{2\pi \: m} \\ \: or \: B = \: \frac{2\pi \: mf}{q} \\ substituting \: \: \: the \: values \: in \: SI \: units.. \: we \: have \\ \: B \: = \frac{(2)( \frac{22}{7})(1.67 \times {10}^{ - 27})(10 \times {10}^{6}) }{1.6 \times {10}^{ - 19} } \\ = 0.67 T \\ \\ kinetic n energy .. ...let the final velocity of proton just after leaving the cyclotron is v . \\ then \: radius \: of \: dee \: should \: be \: equal \: to \: \\ R \: = \frac{mv}{Bq} \: or \: v \: = \frac{BqR}{m} \\ \\ Kinetic \: \:\: energy \:\: of \: proton \: \\ K \: = \frac{1}{2} m {v}^{2} = \frac{1}{2} \times m \times ({\frac{BqR}{m}})^{2} \: = \frac{ {B}^{2} {q}^{2} {R}^{2} }{2m} \\ \\ substituting \: the \: values \: in \: SI \: unit ...we \: have \: \\ \\ k = \frac{ {(0.67)}^{2} {(1.6 \times {10}^{ - 19}) }^{2} {(0.60)}^{2} }{2 \times 1.67 \times {10}^{ - 27} } = 1.2 \times {10}^{ - 12} J \: [/tex]
[tex]f \: = \frac{ Beq }{2\pi \: m} \\ \: or \: B = \: \frac{2\pi \: mf}{q} \\ substituting \: \: \: the \: values \: in \: SI \: units.. \: we \: have \\ \: B \: = \frac{(2)( \frac{22}{7})(1.67 \times {10}^{ - 27})(10 \times {10}^{6}) }{1.6 \times {10}^{ - 19} } \\ = 0.67 T \\ \\ kinetic n energy .. ...let the final velocity of proton just after leaving the cyclotron is v . \\ then \: radius \: of \: dee \: should \: be \: equal \: to \: \\ R \: = \frac{mv}{Bq} \: or \: v \: = \frac{BqR}{m} \\ \\ Kinetic \: \:\: energy \:\: of \: proton \: \\ K \: = \frac{1}{2} m {v}^{2} = \frac{1}{2} \times m \times ({\frac{BqR}{m}})^{2} \: = \frac{ {B}^{2} {q}^{2} {R}^{2} }{2m} \\ \\ substituting \: the \: values \: in \: SI \: unit ...we \: have \: \\ \\ k = \frac{ {(0.67)}^{2} {(1.6 \times {10}^{ - 19}) }^{2} {(0.60)}^{2} }{2 \times 1.67 \times {10}^{ - 27} } = 1.2 \times {10}^{ - 12} J \: [/tex]
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Required magnetic field for accelerating proton
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